Question

The position of an object moving along x-axis is given by x = a + bt² where a = 8.5 m, b = 2.5 m/s² and t is measured in seconds. What is the average velocity between t = 2.0 s and t = 4.0 s ?​

Answer 1

ANSWER

Position is given as

x = a + bt² = 8.5 + 2.5t²

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Position at

t = 2s, ${\mathsf{x_2}}$ = 8.5 + 2.5(2)² = 18.5m

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Position at

t = 4s, ${\mathsf{x_1}}$ = 8.5 + 2.5(4)² = 48.5m

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Displacement

S = ${\mathsf{x_2}}$ - ${\mathsf{x_1}}$ = 48.5 − 18.5 = 30m

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Time taken

t= 4 − 2 = 2s

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Average velocity

${\mathsf{V_{avg}}}$ = ${\mathsf{\frac{S}{t}}}$ = ${\mathsf{\frac{32}{2}}}$ = 15m/s

${\: \:}$${\: \:}$

∵ 15m/s

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Hope this helps ya! ♡

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~Siya♪♫

Answer 2

Explanation:

given x= a+ bt² and a=8.5m and b=2.5m/s²

thus, x=8.5 + 2.5t²

Now,

velocity will be

v= dx/dt

v= d(8.5 + 2.5t²)/dt

v= 5t m/s

Now,

velocity at t= 0s will be, v= 5(0) = 0m/s

velocity at t = 2s will be, v= 5(2) = 10m/s

velocity at t= 4s will be , v= 5(4) = 20m/s

so avg velocity is,

(20+10)/2

= 15m/s